In a recent failed-post about some partial sums with respect to the Central Binomial and Catalan number the formulas
$$\sum_{k=0}^n\frac{4^k}{B_k}=\frac{4^{n+1}(2n+1)}{3 B_{n+1}}+\frac{1}{3}$$
$$\sum_{k=0}^n\frac{4^k(k+1)}{B_k}=\frac{4^{n+1}(2n^2+5n+2)}{5 B_{n+1}}+\frac{1}{5}$$
were mention, here in MO, and I forgot to ask, so let me do it now:

Is This just one instance of some broader well known pattern?

Here, consider $B_m={2m \choose m}$ and $C_m=\frac{B_m}{m+1}$ for those set of numbers.

$\begingroup$you are right, i make a mistake because i should write the next formulas: $\sum_{k=0}^n\frac{4^k}{B_k}=\frac{4^{n+1}(2n+1)}{3B_{n+1}}+\frac{1}{3}$ and $\sum_{k=0}^n\frac{4^k(k+1)}{B_k}=\frac{4^{n+1}(2n^2+5n+2)}{5B_{n+1}}+\frac{1}{5}$ $\endgroup$
– janmarqzJun 18 '11 at 1:04